What is the integral of x/((1+x)^4) ?

The integral of x/((1+x)^4) is -1/(2(1+x)^2)+1/(3(1+x)^3)+C

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integral of x/((1+x)^4)

\int \frac{x}{( 1 + x ) ^{4}} d x

- \frac{1}{2 ( 1 + x ) ^{2}} + \frac{1}{3 ( 1 + x ) ^{3}} + C

Solution steps

\int \frac{x}{( 1 + x ) ^{4}} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{4}} d u

= \int \frac{u - 1}{u ^{4}} d u

Expand

\frac{u - 1}{u ^{4}} : \frac{1}{u ^{3}} - \frac{1}{u ^{4}}

= \int \frac{1}{u ^{3}} - \frac{1}{u ^{4}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{3}} d u - \int \frac{1}{u ^{4}} d u

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

= - \frac{1}{2 u ^{2}} - (- \frac{1}{3 u ^{3}})

Substitute back

u = 1 + x

= - \frac{1}{2 ( 1 + x ) ^{2}} - (- \frac{1}{3 ( 1 + x ) ^{3}})

Simplify

= - \frac{1}{2 ( 1 + x ) ^{2}} + \frac{1}{3 ( 1 + x ) ^{3}}

Add a constant to the solution

= - \frac{1}{2 ( 1 + x ) ^{2}} + \frac{1}{3 ( 1 + x ) ^{3}} + C

\int x (x^{4} + \frac{1}{x ^{2}}) d x

\int (t^{2} + t) e^{t^{3} + 3 t} d x

y^{^{'}} + 3 y = 2

\int_{1}^{2} (\frac{2}{x ^{3}}) d x

f (x) = ln (x^{3} - 1)

What is the integral of x/((1+x)^4) ?
The integral of x/((1+x)^4) is -1/(2(1+x)^2)+1/(3(1+x)^3)+C